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(C) The velocity of the first particle is given by the derivative of its position with respect to time: $v_x = \frac{dx}{dt} = 8 - 6t$. At $t = 1 \; s

Vedclass · Accepted Answer

$580$

Vedclass · Answer

(C) The velocity of the first particle is given by the derivative of its position with respect to time: $v_x = \frac{dx}{dt} = 8 - 6t$. At $t = 1 \; s$,$v_x = 8 - 6(1) = 2 \; m/s$. Thus,$\vec{v}_1 = 2 \hat{i} \; m/s$.The velocity of the second particle is given by the derivative of its position with respect to time: $v_y = \frac{dy}{dt} = -24t^2$. At $t = 1 \; s$,$v_y = -24(1)^2 = -24 \; m/s$. Thus,$\vec{v}_2 = -24 \hat{j} \; m/s$.The velocity of the second particle relative to the first particle is $\vec{v}_{21} = \vec{v}_2 - \vec{v}_1 = -2 \hat{i} - 24 \hat{j}$.The speed of the second particle in the frame of the first is the magnitude of the relative velocity: $|\vec{v}_{21}| = \sqrt{(-2)^2 + (-24)^2} = \sqrt{4 + 576} = \sqrt{580}$.Given that the speed is $\sqrt{v}$,we have $\sqrt{v} = \sqrt{580}$,which implies $v = 580$.

Column-$I$	Column-$II$
$(1)$ Velocity of $A$ relative to $B$ if $A$ and $B$ are moving perpendicular to each other	$(a)$ $\vec{v}_{rm} = \vec{v}_r + \vec{v}_m$
$(2)$ Velocity of rain drops relative to a man	$(b)$ $\vec{v}_{AB} = \vec{v}_A + \vec{v}_B$
	$(c)$ $v_{AB} = \sqrt{v_A^2 + v_B^2}$

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